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(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$.The angle b

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(D) Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,so $|\overrightarrow{a}| = 1$ and $|\overrightarrow{b}| = 1$.The angle between them is $	heta = \frac{\pi}{3}$,so $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos(\frac{\pi}{3}) = 1 	imes 1 	imes \frac{1}{2} = \frac{1}{2}$.Since the vectors $(\lambda \overrightarrow{a} + 2 \overrightarrow{b})$ and $(3 \overrightarrow{a} - \lambda \overrightarrow{b})$ are perpendicular,their dot product is $0$:$(\lambda \overrightarrow{a} + 2 \overrightarrow{b}) \cdot (3 \overrightarrow{a} - \lambda \overrightarrow{b}) = 0$$3\lambda (\overrightarrow{a} \cdot \overrightarrow{a}) - \lambda^2 (\overrightarrow{a} \cdot \overrightarrow{b}) + 6 (\overrightarrow{a} \cdot \overrightarrow{b}) - 2\lambda (\overrightarrow{b} \cdot \overrightarrow{b}) = 0$Since $\overrightarrow{a} \cdot \overrightarrow{a} = 1$ and $\overrightarrow{b} \cdot \overrightarrow{b} = 1$,we have:$3\lambda - \lambda^2(\frac{1}{2}) + 6(\frac{1}{2}) - 2\lambda = 0$$3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = 0$$\lambda - \frac{\lambda^2}{2} + 3 = 0$Multiplying by $-2$,we get $\lambda^2 - 2\lambda - 6 = 0$.The roots are $\lambda = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$.Since $\sqrt{7} \approx 2.64$,the values are $\lambda_1 = 1 + 2.64 = 3.64$ and $\lambda_2 = 1 - 2.64 = -1.64$.Neither $3.64$ nor $-1.64$ lies in the interval $[-1, 3]$.Thus,the number of values of $\lambda$ in $[-1, 3]$ is $0$.

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